fractional derivaitve of logarithm function $x^ {a} \log(x) $

Question

Given the function $ x^{a}\log(x) $ natural logarithmic

Could someone tell me how to evaluate the fractional derivative

$$ \frac{d^{b}}{dx^{b}}x^{a}\log(x) $$ for positive $a$ and $b$

Which fractional derivative are you considering? The main difficulty here is that the Leibniz rule doesn't hold for fractional-order derivatives. There are however certain extensions. For the Riemann-Liouville case, check out Section 2.7.2 in: I. Podlubny, "Fractional Differential Equations," Academic Press, 1999. For the Caputo derivative see Eq. (9) in K. Diethelm et al., "Algorithms for the fractional calculus: A selection of numerical methods," Comp. meth. appl. mech. eng. 194 (2005). — Pantelis Sopasakis, Dec 10 '14 at 11:15

score 2 · Accepted Answer · answered Jun 17 '16 at 15:15

There is an extension to the fractional derivative for the General Leibniz rule:

$$(uv)^{(n)}=\sum_{k=0}^\infty\frac{\Gamma(n+1)}{k!\Gamma(n-k+1)}u^{(n-k)}v^{(k)}$$

For your case,

$$(x^a\ln(x))^{(b)}=\frac{\Gamma(a+1)}{\Gamma(a-b+1)}x^{a-b}\ln(x)+\sum_{k=1}^\infty\frac{\Gamma(b+1)}{k!\Gamma(b-k+1)}(x^a)^{(b-k)}\ln(x)^{(k)}$$

$$=\frac{\Gamma(a+1)}{\Gamma(a-b+1)}x^{a-b}\ln(x)+\sum_{k=1}^\infty\frac{(-1)^{k+1}\Gamma(b+1)\Gamma(a+1)}{\Gamma(b-k+1)\Gamma(a+k-b+1)}x^{a-b}$$

Of course, this probably isn't the only extension to the Leibniz rule, as Pantelis Sopasakis explains. — Simply Beautiful Art, Jun 17 '16 at 18:44

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